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Grade upto college level Physical Chemistry

0.5 gm of fuming H2SO4 (Oleum) is dilute with water. This solution is completely neutralized by 26.7 ml of 0.4 N NaOH. Find the percentage of free SO3 in the sample of oleum.

Profile image of Shane Macguire
12 Years agoGrade upto college level
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1 Answer

Profile image of Deepak Patra
12 Years ago
Hello Student,
Please find the answer to your question
N1= I, Vl=?,N2=26.7, V2=0.4
N1v1'= N2v2
1 x V1 =26.7 x 0.4
V1 = 26.7 * 0.4/1 = 10.68
49g (∵ eqwt of H2SO4 = 49) pf H2SO4 will be neutralized by = 1N 1000 ml NaOH
∴ 0.5g of H2SO4 will be neutralized
= 1000/49 * 0.5 = 10.20 ml 1N NaOH
Volume of 1 N NaOH used by dissolved
SO3 = 10.68 – 10.20 = 0.48 ml
SO3 + 2NaOH → Na2SO4 + H2O
∴ Eqwt of SO3 = Molwt/2 = 80/2 = 40
Wt of SO3 in 0.48 ml of 1 M solution
= 40/1000 * 0.48 = 0.0192 g
% of SO3 = 0.0192/0.5 * 100 = 3.84 %

Thanks
Deepak patra
askIITians Faculty